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a. The continuous growth rate of the bacteria is 21%
b. The initial population of bacteria is 715
c. The culture will contain 2043 bacteria after 6 × 10⁻⁴ years
<h3>a. What is the continuous rate of growth of this bacteria population?</h3>Since represents the number of bacteria in the culture.
This function is similar to an exponential function of the form where λ = growth rate
Comparing n(t) and y(t), we see that λ = 0.21
So, the continuous growth rate of the bacteria is 0.21 = 0.21 × 100 %
= 21%
So, the continuous growth rate of the bacteria is 21%
<h3>b. What is the initial population of the culture?</h3>Since represents the number of bacteria in the culture, the initial population of bacteria is obtained when t = 0.
So,
So, the initial population of bacteria is 715
<h3>c. When will the culture contain 2043 bacteria?</h3>To find the time when the number of bacteria will be 2043, this means n(t) = 2043.
Since
Making t subject of the formula, we have
t = ㏑[n(t)/715]/0.21
So, substituting n(t) = 2043 into the equation, we have
t = ㏑[n(t)/715]/0.21
t = ㏑[2043/715]/0.21
t = ㏑[2.8573]/0.21
t = 1.05/0.21
t = 4.99
t ≅ 5 hours
Converting this to years, we have t = 5 h × 1 day/24h × 1 year/365 days
= 5/8760
= 5.7 × 10⁻⁴ years
≅ 6 × 10⁻⁴ years
So, the culture will contain 2043 bacteria after 6 × 10⁻⁴ years
Learn more about exponential bacteria growth here:
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